\(\int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+(e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))) \log (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5})}{e^5 (3 x^3+x^4)+e^5 x^3 \log (x)+(3 x^3+x^4+x^3 \log (x)) \log (-3-x-\log (x))} \, dx\) [646]

Optimal result

Integrand size = 145, antiderivative size = 22 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\log \left (x+\frac {x \log (-3-x-\log (x))}{e^5}\right )}{x^2} \] Output:

ln(x+ln(-ln(x)-3-x)*x/exp(5))/x^2
 

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\log \left (x+\frac {x \log (-3-x-\log (x))}{e^5}\right )}{x^2} \] Input:

Integrate[(1 + x + E^5*(3 + x) + E^5*Log[x] + (3 + x + Log[x])*Log[-3 - x 
- Log[x]] + (E^5*(-6 - 2*x) - 2*E^5*Log[x] + (-6 - 2*x - 2*Log[x])*Log[-3 
- x - Log[x]])*Log[(E^5*x + x*Log[-3 - x - Log[x]])/E^5])/(E^5*(3*x^3 + x^ 
4) + E^5*x^3*Log[x] + (3*x^3 + x^4 + x^3*Log[x])*Log[-3 - x - Log[x]]),x]
 

Output:

Log[x + (x*Log[-3 - x - Log[x]])/E^5]/x^2
 

Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x+e^5 (x+3)+e^5 \log (x)+(x+\log (x)+3) \log (-x-\log (x)-3)+\left (e^5 (-2 x-6)-2 e^5 \log (x)+(-2 x-2 \log (x)-6) \log (-x-\log (x)-3)\right ) \log \left (\frac {e^5 x+x \log (-x-\log (x)-3)}{e^5}\right )+1}{e^5 x^3 \log (x)+e^5 \left (x^4+3 x^3\right )+\left (x^4+3 x^3+x^3 \log (x)\right ) \log (-x-\log (x)-3)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {x+e^5 (x+3)+e^5 \log (x)+(x+\log (x)+3) \log (-x-\log (x)-3)+\left (e^5 (-2 x-6)-2 e^5 \log (x)+(-2 x-2 \log (x)-6) \log (-x-\log (x)-3)\right ) \log \left (\frac {e^5 x+x \log (-x-\log (x)-3)}{e^5}\right )+1}{x^3 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^5 (x+3)}{x^3 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}-\frac {2 \log \left (\frac {x \left (\log (-x-\log (x)-3)+e^5\right )}{e^5}\right )}{x^3}+\frac {\log (-x-\log (x)-3)}{x^3 \left (\log (-x-\log (x)-3)+e^5\right )}+\frac {e^5 \log (x)}{x^3 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}+\frac {1}{x^3 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}+\frac {1}{x^2 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -e^5 \int \frac {1}{x^3 \left (\log (-x-\log (x)-3)+e^5\right )}dx+3 e^5 \int \frac {1}{x^3 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}dx+e^5 \int \frac {\log (x)}{x^3 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}dx+e^5 \int \frac {1}{x^2 (x+\log (x)+3) \left (\log (-x-\log (x)-3)+e^5\right )}dx+\frac {\log \left (\frac {x \left (\log (-x-\log (x)-3)+e^5\right )}{e^5}\right )}{x^2}\)

Input:

Int[(1 + x + E^5*(3 + x) + E^5*Log[x] + (3 + x + Log[x])*Log[-3 - x - Log[ 
x]] + (E^5*(-6 - 2*x) - 2*E^5*Log[x] + (-6 - 2*x - 2*Log[x])*Log[-3 - x - 
Log[x]])*Log[(E^5*x + x*Log[-3 - x - Log[x]])/E^5])/(E^5*(3*x^3 + x^4) + E 
^5*x^3*Log[x] + (3*x^3 + x^4 + x^3*Log[x])*Log[-3 - x - Log[x]]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 16.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {\ln \left (x \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right ) {\mathrm e}^{-5}\right )}{x^{2}}\)	\(25\)
risch	\(\frac {\ln \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )}{x^{2}}+\frac {i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )\right ) \operatorname {csgn}\left (i x \right )-10+i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )\right )}^{2} \operatorname {csgn}\left (i x \right )-i \pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{5}+\ln \left (-\ln \left (x \right )-3-x \right )\right )\right )}^{3}+2 \ln \left (x \right )}{2 x^{2}}\)	\(169\)

Input:

int((((-2*ln(x)-2*x-6)*ln(-ln(x)-3-x)-2*exp(5)*ln(x)+(-2*x-6)*exp(5))*ln(( 
x*ln(-ln(x)-3-x)+x*exp(5))/exp(5))+(3+x+ln(x))*ln(-ln(x)-3-x)+exp(5)*ln(x) 
+(3+x)*exp(5)+x+1)/((x^3*ln(x)+x^4+3*x^3)*ln(-ln(x)-3-x)+x^3*exp(5)*ln(x)+ 
(x^4+3*x^3)*exp(5)),x,method=_RETURNVERBOSE)
 

Output:

1/x^2*ln(x*(exp(5)+ln(-ln(x)-3-x))/exp(5))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\log \left ({\left (x e^{5} + x \log \left (-x - \log \left (x\right ) - 3\right )\right )} e^{\left (-5\right )}\right )}{x^{2}} \] Input:

integrate((((-2*log(x)-2*x-6)*log(-log(x)-3-x)-2*exp(5)*log(x)+(-2*x-6)*ex 
p(5))*log((x*log(-log(x)-3-x)+x*exp(5))/exp(5))+(3+x+log(x))*log(-log(x)-3 
-x)+exp(5)*log(x)+(3+x)*exp(5)+x+1)/((x^3*log(x)+x^4+3*x^3)*log(-log(x)-3- 
x)+x^3*exp(5)*log(x)+(x^4+3*x^3)*exp(5)),x, algorithm="fricas")
 

Output:

log((x*e^5 + x*log(-x - log(x) - 3))*e^(-5))/x^2
 

Sympy [A] (verification not implemented)

Time = 3.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\log {\left (\frac {x \log {\left (- x - \log {\left (x \right )} - 3 \right )} + x e^{5}}{e^{5}} \right )}}{x^{2}} \] Input:

integrate((((-2*ln(x)-2*x-6)*ln(-ln(x)-3-x)-2*exp(5)*ln(x)+(-2*x-6)*exp(5) 
)*ln((x*ln(-ln(x)-3-x)+x*exp(5))/exp(5))+(3+x+ln(x))*ln(-ln(x)-3-x)+exp(5) 
*ln(x)+(3+x)*exp(5)+x+1)/((x**3*ln(x)+x**4+3*x**3)*ln(-ln(x)-3-x)+x**3*exp 
(5)*ln(x)+(x**4+3*x**3)*exp(5)),x)
 

Output:

log((x*log(-x - log(x) - 3) + x*exp(5))*exp(-5))/x**2
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\log \left (x\right ) + \log \left (e^{5} + \log \left (-x - \log \left (x\right ) - 3\right )\right ) - 5}{x^{2}} \] Input:

integrate((((-2*log(x)-2*x-6)*log(-log(x)-3-x)-2*exp(5)*log(x)+(-2*x-6)*ex 
p(5))*log((x*log(-log(x)-3-x)+x*exp(5))/exp(5))+(3+x+log(x))*log(-log(x)-3 
-x)+exp(5)*log(x)+(3+x)*exp(5)+x+1)/((x^3*log(x)+x^4+3*x^3)*log(-log(x)-3- 
x)+x^3*exp(5)*log(x)+(x^4+3*x^3)*exp(5)),x, algorithm="maxima")
 

Output:

(log(x) + log(e^5 + log(-x - log(x) - 3)) - 5)/x^2
 

Giac [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\log \left (x\right ) + \log \left (e^{5} + \log \left (-x - \log \left (x\right ) - 3\right )\right ) - 5}{x^{2}} \] Input:

integrate((((-2*log(x)-2*x-6)*log(-log(x)-3-x)-2*exp(5)*log(x)+(-2*x-6)*ex 
p(5))*log((x*log(-log(x)-3-x)+x*exp(5))/exp(5))+(3+x+log(x))*log(-log(x)-3 
-x)+exp(5)*log(x)+(3+x)*exp(5)+x+1)/((x^3*log(x)+x^4+3*x^3)*log(-log(x)-3- 
x)+x^3*exp(5)*log(x)+(x^4+3*x^3)*exp(5)),x, algorithm="giac")
 

Output:

(log(x) + log(e^5 + log(-x - log(x) - 3)) - 5)/x^2
 

Mupad [B] (verification not implemented)

Time = 3.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\ln \left (x\,\left ({\mathrm {e}}^5+\ln \left (-x-\ln \left (x\right )-3\right )\right )\right )-5}{x^2} \] Input:

int((x + log(- x - log(x) - 3)*(x + log(x) + 3) + exp(5)*(x + 3) + exp(5)* 
log(x) - log(exp(-5)*(x*exp(5) + x*log(- x - log(x) - 3)))*(2*exp(5)*log(x 
) + log(- x - log(x) - 3)*(2*x + 2*log(x) + 6) + exp(5)*(2*x + 6)) + 1)/(e 
xp(5)*(3*x^3 + x^4) + log(- x - log(x) - 3)*(x^3*log(x) + 3*x^3 + x^4) + x 
^3*exp(5)*log(x)),x)
 

Output:

(log(x*(exp(5) + log(- x - log(x) - 3))) - 5)/x^2
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {1+x+e^5 (3+x)+e^5 \log (x)+(3+x+\log (x)) \log (-3-x-\log (x))+\left (e^5 (-6-2 x)-2 e^5 \log (x)+(-6-2 x-2 \log (x)) \log (-3-x-\log (x))\right ) \log \left (\frac {e^5 x+x \log (-3-x-\log (x))}{e^5}\right )}{e^5 \left (3 x^3+x^4\right )+e^5 x^3 \log (x)+\left (3 x^3+x^4+x^3 \log (x)\right ) \log (-3-x-\log (x))} \, dx=\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (-\mathrm {log}\left (x \right )-x -3\right ) x +e^{5} x}{e^{5}}\right )}{x^{2}} \] Input:

int((((-2*log(x)-2*x-6)*log(-log(x)-3-x)-2*exp(5)*log(x)+(-2*x-6)*exp(5))* 
log((x*log(-log(x)-3-x)+x*exp(5))/exp(5))+(3+x+log(x))*log(-log(x)-3-x)+ex 
p(5)*log(x)+(3+x)*exp(5)+x+1)/((x^3*log(x)+x^4+3*x^3)*log(-log(x)-3-x)+x^3 
*exp(5)*log(x)+(x^4+3*x^3)*exp(5)),x)
 

Output:

log((log( - log(x) - x - 3)*x + e**5*x)/e**5)/x**2